# Thread: Series dealing with exponationals

1. ## Series dealing with exponationals

I don't see how

1. $\displaystyle \sum_{n=1}^\infty \frac{1}{e^n} + \frac{1}{n(n+1)}$ = $\displaystyle \frac{e}{e-1}$

2. $\displaystyle \sum_{n=1}^\infty e^\frac{1}{n} - e^\frac{1}{n+1}$ = $\displaystyle e-1$

In the first case I know that as $\displaystyle n\rightarrow\infty$ that $\displaystyle \frac{1}{e^n}$ approaches 0 and therefore it converges, and I know that $\displaystyle \frac{1}{n(n+1)} = 1$ (from a previous problem). But how do you know what $\displaystyle \frac{1}{e^n}$ sums up to?.

And in the second case I have no clue about approaching the problem, but I would think that it's the same type of approach as the first.

2. $\displaystyle \sum\limits_{n = 1}^\infty {\left( {\frac{1} {e}} \right)^n }$

is just an infinite geometric series, and we know that it converges since 1/e < 1 thus:

$\displaystyle \sum\limits_{n = 1}^\infty {\left( {\frac{1} {e}} \right)^n } = \frac{{a_1 }} {{1 - q}} = \frac{{\frac{1} {e}}} {{1 - \frac{1} {e}}} = \frac{1} {{e - 1}}$

the second term can be rewritten using partial fraction decomposition as:

$\displaystyle \sum\limits_{n = 1}^\infty {\frac{1} {{n\left( {n + 1} \right)}} = } \sum\limits_{n = 1}^\infty {\frac{1} {n}} - \frac{1} {{n + 1}}$

$\displaystyle \sum\limits_{n = 1}^\infty {\frac{1} {n}} - \frac{1} {{n + 1}} = \left( {1 - \frac{1} {2}} \right) + \left( {\frac{1} {2} - \frac{1} {3}} \right) + \left( {\frac{1} {3} - \frac{1} {4}} \right) + \cdots = 1$

since both individual terms converge to 0.
so:

$\displaystyle \sum\limits_{n = 1}^\infty {\frac{1} {{e^n }}} + \sum\limits_{n = 1}^\infty {\frac{1} {{n\left( {n + 1} \right)}} = \frac{1} {{e - 1}} + 1 = \frac{e} {{e - 1}}}$

$\displaystyle \sum\limits_{n = 1}^\infty {e^{\frac{1} {n}} - e^{\frac{1} {{n + 1}}} = \left( {e - e^{\frac{1} {2}} } \right) + \left( {e^{\frac{1} {2}} - e^{\frac{1} {3}} } \right) + \cdots }$

now let's look at the sum of the first N elements and then take the limit as N approaches infinity:

$\displaystyle \mathop {\lim }\limits_{N \to \infty } \sum\limits_{n = 1}^N {e^{\frac{1} {n}} - e^{\frac{1} {{n + 1}}} = \left( {e - e^{\frac{1} {2}} } \right) + \left( {e^{\frac{1} {2}} - e^{\frac{1} {3}} } \right) + \cdots }$$\displaystyle + \left( {e^{\frac{1} {{N - 1}}} - e^{\frac{1} {N}} } \right) + \left( {e^{\frac{1} {N}} - e^{\frac{1} {{N + 1}}} } \right) = e - 1$